I get this code for LARS but the variable seems undefined? - matlab

Question

I got this code for LARS but when I run, it says undefined X. I can't understand what x is. Why is there an error?
function [beta, A, mu, C, c, gamma] = lars(X, Y, option, t, standardize)
% Least Angle Regression (LAR) algorithm.
% Ref: Efron et. al. (2004) Least angle regression. Annals of Statistics.
% option = 'lar' implements the vanilla LAR algorithm (default);
% option = 'lasso' solves the lasso path with a modified LAR algorithm.
% t -- a vector of increasing positive real numbers. If given, LARS
% returns the solution at t.
%
% Output:
% A -- a sequence of indices that indicate the order of variable
% beta: history of estimated LARS coefficients;
% mu -- history of estimated mean vector;
% C -- history of maximal current absolute corrrelations;
% c -- history of current corrrelations;
% gamma: history of LARS step size.
% Note: history is traced by rows. If t is given, beta is just the
% estimated coefficient vector at the constraint ||beta||_1 = t.
%
% Remarks:
% 1. LARS is originally proposed to estimate a sparse coefficient vector
% a noisy over-determined linear system. LARS outputs estimates for all
% shrinkage/constraint parameters (homotopy).
%
% 2. LARS is well suited for Basis Pursuit (BP) purpose in the real
% automatically terminates when the current correlations for inactive
% all zeros. The recovered coefficient vector is the last column of beta
% with the *lasso* option. Hence, this function provides a fast and
% efficient solution for the ell_1 minimization problem.
% Ref: Donoho and Tsaig (2006). Fast solution of ell_1 norm minimization
if nargin < 5, standardize = true; end
if nargin < 4, t = Inf; end
if nargin < 3, option = 'lar'; end
if strcmpi(option, 'lasso'), lasso = 1; else, lasso = 0; end
eps = 1e-10; % Effective zero
[n,p] = size(X);
if standardize,
X = normalize(X);
Y = Y-mean(Y);
end
m = min(p,n-1); % Maximal number of variables in the final active set
T = length(t);
beta = zeros(1,p);
mu = zeros(n,1); % Mean vector
gamma = []; % LARS step lengths
A = [];
Ac = 1:p;
nVars = 0;
signOK = 1;
i = 0;
mu_old = zeros(n,1);
t_prev = 0;
beta_t = zeros(T,p);
ii = 1;
tt = t;
% LARS loop
while nVars < m,
i = i+1;
c = X'*(Y-mu); % Current correlation
C = max(abs(c)); % Maximal current absolute correlation
if C < eps || isempty(t), break; end % Early stopping criteria
if 1 == i, addVar = find(C==abs(c)); end
if signOK,
A = [A,addVar]; % Add one variable to active set
nVars = nVars+1;
end
s_A = sign(c(A));
Ac = setdiff(1:p,A); % Inactive set
nZeros = length(Ac);
X_A = X(:,A);
G_A = X_A'*X_A; % Gram matrix
invG_A = inv(G_A);
L_A = 1/sqrt(s_A'*invG_A*s_A);
w_A = L_A*invG_A*s_A; % Coefficients of equiangular vector u_A
u_A = X_A*w_A; % Equiangular vector
a = X'*u_A; % Angles between x_j and u_A
beta_tmp = zeros(p,1);
gammaTest = zeros(nZeros,2);
if nVars == m,
gamma(i) = C/L_A; % Move to the least squares projection
else
for j = 1:nZeros,
jj = Ac(j);
gammaTest(j,:) = [(C-c(jj))/(L_A-a(jj)), (C+c(jj))/(L_A+a(jj))];
end
[gamma(i) min_i min_j] = minplus(gammaTest);
addVar = unique(Ac(min_i));
end
beta_tmp(A) = beta(i,A)' + gamma(i)*w_A; % Update coefficient estimates
% Check the sign feasibility of lasso
if lasso,
signOK = 1;
gammaTest = -beta(i,A)'./w_A;
[gamma2 min_i min_j] = minplus(gammaTest);
if gamma2 < gamma(i), % The case when sign consistency gets violated
gamma(i) = gamma2;
beta_tmp(A) = beta(i,A)' + gamma(i)*w_A; % Correct the coefficients
beta_tmp(A(unique(min_i))) = 0;
A(unique(min_i)) = []; % Delete the zero-crossing variable (keep the ordering)
nVars = nVars-1;
signOK = 0;
end
end
if Inf ~= t(1),
t_now = norm(beta_tmp(A),1);
if t_prev < t(1) && t_now >= t(1),
beta_t(ii,A) = beta(i,A) + L_A*(t(1)-t_prev)*w_A'; % Compute coefficient estimates corresponding to a specific t
t(1) = [];
ii = ii+1;
end
t_prev = t_now;
end
mu = mu_old + gamma(i)*u_A; % Update mean vector
mu_old = mu;
beta = [beta; beta_tmp'];
end
if 1 < ii,
noCons = (tt > norm(beta_tmp,1));
if 0 < sum(noCons),
beta_t(noCons,:) = repmat(beta_tmp',sum(noCons),1);
end
beta = beta_t;
end
% Normalize columns of X to have mean zero and length one.
function sX = normalize(X)
[n,p] = size(X);
sX = X-repmat(mean(X),n,1);
sX = sX*diag(1./sqrt(ones(1,n)*sX.^2));
% Find the minimum and its index over the (strictly) positive part of X
% matrix
function [m, I, J] = minplus(X)
% Remove complex elements and reset to Inf
[I,J] = find(0~=imag(X));
for i = 1:length(I),
X(I(i),J(i)) = Inf;
end
X(X<=0) = Inf;
m = min(min(X));
[I,J] = find(X==m);

Answer 1

You can have more information in the related paper:
Efron, Bradley; Hastie, Trevor; Johnstone, Iain; Tibshirani, Robert. Least angle regression. Ann. Statist. 32 (2004), no. 2, 407--499. doi:10.1214/009053604000000067.
http://projecteuclid.org/euclid.aos/1083178935.